GTU CHEM 113 GENERAL CHEMISTRY LABORATORY-I BOOKLET

2016 FALL

PREFACE Üniversitelerde Fen bölümleri, tıp ve mühendislik öğrencileri ilk yıllarında Kimya Laboratuvarı dersi ile karşılaşmaktadır. Kimya Laboratuvarı dersi ile ilgili bu kitapçığın başında öncelikle kimya laboratuvarı için son derece önemli olan kimya laboratuvarında uyulacak kurallar ve laboratuvarda alınması gereken güvenlik önlemleri açıklanmıştır. Laboratuvar güvenliği ile ilgili detaylı bilgiler, verilen web bağlantısındaki kaynakta yer almaktadır.

Kimya Laboratuvarında kullanılan malzemelerin tanıtıldığı bölümden sonra,

Genel Kimya Ders konuları ile bağlantılı seçilen deneylerin föyleri yer almaktadır. Laboratuvara ilk geldiğiniz gün, kimya laboratuvarında uyulacak kurallar ve laboratuvarda alınması gereken güvenlik önlemleri konularında bilgilendirilerek, yangın tüpleri, güvenlik duşu, göz yıkama muslukları gibi kaza anında kullanılacak malzeme ve sistemlerin yerleri ve nasıl çalıştıklarını öğreneceksiniz. Kimya laboratuvarları, her an istenmeyen kazaların yaşanabileceği yerler olması nedeni ile herşeyden önce güvenlik konularındaki bilgilerin öğrenilmesi çok önemlidir.

İlk hafta

güvenlik kuralları ve önlemleri bölümünü ve her deney haftası deney föyünü mutlaka okuyup öğrenerek geliniz. Hepinize sağlıklı, başarılı ve kazasız bir eğitim, öğretim dönemi dilerim.

Saygılarımla Prof. Dr. Ayşe Gül GÜREK Kimya Bölüm Başkan V.

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CONTENT

PREFACE .................................................................................................................................. 2 CONTENT ................................................................................................................................. 3 LABORATORY RULES FOR CHEM 113 GENERAL CHEMISTRY LAB. I COURSE ...... 4 LABORATORY EQUIPMENTS ............................................................................................ 11 EXPERIMENT 1: IDENTIFICATION OF PURE SUBSTANCES BY THEIR PROPERTIES .................................................................................................................................................. 16 EXPERIMENT 2: IDENTIFICATION OF SUBSTANCES BY THEIR REACTIONS………………………………………………………………………………...20 EXPERIMENT 3: STOCHIOMETRIC CALCULATIONS ................................................... 26 EXPERIMENT 4: ACID - BASE TITRATION ..................................................................... 29 EXPERIMENT 5: DETERMINATION OF A SALT SOLUBILITY ..................................... 36 EXPERIMENT 6: DETERMINATION QUANTITY OF Fe2+ WITH KMNO4 ..................... 40 EXPERIMENT 7: BUFFERS, BUFFER CAPACITY, AND BUFFERING ZONE .............. 44 EXPERIMENT 8: POLYPROTIC ACIDS: DETERMINATION OF pKA VALUES USING pH TITRATION CURVES ...................................................................................................... 53

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LABORATORY RULES FOR CHEM 113 GENERAL CHEMISTRY LAB. I COURSE 1.

The students must attend the General Chemistry Laboratory course in exact day, time and

laboratory, which are announced on the course programs’ of the departments. 2.

Attendance compulsion is %80 for the laboratory semester. If a student does not attend for at

least 2 experiments, a VF will be issued. 3.

During the lab course, the students have to wear their lab coat, lab goggles and latex gloves.

The students must bring their own lab coat, lab goggles and latex gloves to the lab. Be aware that failing to wear lab coat, lab goggles and latex gloves in the laboratory will result in immediate expulsion from the laboratory, failing the experiment. This rule is strictly enforced. 4.

If the student is late more than 10 minutes to the lab course, she/he will not to be allowed to

attend the lab and the experiment. 5.

During the lab course, it is strictly forbidden to use mobile phones. The students must not

leave the lab and experiment set without permission of the lab assistant. 6.

In order to provide life safety and to ensure the success of experiment in a safe manner, hand

jokes among the students are strictly forbidden. Also, touching to chemicals without gloves, sniffing and tasting them are dangerous. 7.

In each experiment, required chemicals and materials will be supplied to the students by lab

assistants. The students will not borrow the chemicals and materials from the other students. At the end of the each experiment, all the glass or metal materials will be cleaned by the students. Also, the experiment set will be tidy and clean. 8.

Before coming to the lab, the students are expected to read the experiment of the week in this

booklet thoroughly and to know about the procedure of the regarding experiment and theoretical information. 9.

At the start of each laboratory, the students will be given a prelab quiz. The object of this quiz

is to assess whether they are well prepared for the experiment. 10.

The total grade of an experiment will be the sum of the report grade (%50) and prelab quiz

grade (%50). 11.

At the beginning of the semester, the students will be informed about the safety rules of

laboratory. After that, the students will be required to pass the written examination about the safety rules in two attempts. If the student does not get 60 out of 100 from the written examination in the first attempt, which will be held at the2nd week of semester, the student will have the second right for the 4

examination as a make-up exam in the 3rd week of semester. Failing to get the score over 60 in both exams will result in failing the General Chemistry Laboratory course. 12.

The final grade of General Chemistry Laboratory course will be based on the sum of the

laboratory safety rules exam (%20) and the average grade of the all experiments for the semester (%80). 13.

Students who do not obey the rules and fulfill the requirements of the lab will be considered as

not attended to the lab. (look at entries 1, 3, 4, 11) and no reports for the mentioned lab session will be accepted. 14.

Make-up experiments will be available only for those with health reports. The health reports

must be handed over within 5 days to the lab assistant.

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LABORATORY SAFETY RULES AND CAUTIONS

1. Lab coats must be worn at all times when in the laboratory! All students have to wear lab coat and students without lab coat will not be allowed in the lab. 2. Safety glasses must be worn at all times when in the laboratory! All students have to wear safety glasses in the lab and students without safety glasses will not be allowed in the lab. Contact lenses are not allowed in the lab because acid and organic chemical vapors could get in between eye and lenses, lenses can be glued on the eye in the case of an accident and can be diffıcult to remove. 3. Dress appropriately! Wide dresses, sandal type shoes shouldn’t be worn in the lab and long hair should be tied. 4. No Foods or Drinks! Food, drink and chewing gum are not allowed in the lab. 5. Know the location of exits and safety equipment! All students should know where the fire extinguisher, first aid cabinet and shower are placed in the lab. 6. Learn the speediest exit from the lab. in the case of a fire. 7. Use the shower when your dresses or hair catches fire. 8. Do not run and do not make jokes in the chemistry lab. 9. Benches should not be used to sit on or to leave bags or personal things. 10. Smoke and vapor released during chemical reactions should not be exposed directly. 11. Do not attempt unauthorized experiments! It is not allowed to work in chemistry labs. without attendance of assistants or instructors. 12. Do not use Bunsen burners next to flammable chemicals (e.g. ethers) 13. Read the labels carefully on the bottle before use of any chemicals. 14. Read the experimental procedure before coming to the lab. Students who come to lab having no knowledge about experiment could create risks for themselves and other students. 15. In the case of any accident (glass cut, acid/base burn, fainting etc.), know where to get help fast and immediately inform your assistant or instructor. 16. Do not orientate the test tube toward yourself and your friend. Reaction carried out in the test tube could be dangerous.

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17. Water should not be added on concentrated acids. Acids should be added to the water slowly and by stirring. 18. It is forbidden to smell, to taste the chemicals or to pull solutions by mouth when using pipet. Immediately clean up any chemicals that you spill. If necessary, obtain advice on the cleanup procedure from the lab assistant. 19. Chemicals (solid, liquid or solution) must not be dumped into the sink. Waste bottles in the lab should be used. (Learn where the waste bottles in the lab are.) 20. Use the ‘’ broken glass labeled container in the lab for the broken glass pieces. 21. Matchstick, litmus paper must not be disposed to the sink. 22. Mercury vapor is invisible and toxic. Mercury in the broken thermometer is very dangerous. You must inform your assistant in the case of thermometer breakage. 23. Materials like hot test tube, crucible, and beaker must not hold by hand. Theyshould be cautiously held by tube tongs or left to cool on an asbestos wire. 24. Please use the amounts of chemicals given in the procedure. Use of excess amounts can make difficult to control the reactions or cause the side reactions. 25. The chemicals left behind the experiment should not be returned to the stock bottles, should be discarded into the waste bottle. 26. Always keep your working area, the balance and its environment clean. Work clean and tidy in the lab. 27. Don’t change the locations of the stock chemicals during the experiment. 28. At the end of the experiment, hand in all the materials you used to your assistant as cleaned. 29. Before leaving the lab, make sure that gas and water taps are closed (turned off). 30. Wash your hands before leaving the lab. 31. Learn the information given in the link below and Table 1. http://www.gtu.edu.tr/Files/kimyaBolumu/documents/LabGuvenlik.pdf Note: This web page is important for Lab. Security Examination General laboratuvary safety exam. covers the information given in the link above (slides ; 132, 40-47 , 52 ) and Table 1.

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The following cases, must be notified firstly to the assistant or lecturer. BURN: expose the burned area to the tap water (5-10 CUT / INJURY: Wash with water min.), apply first aid. and apply first aid. FAINTING: Provide fresh air. Lay down and put the FIRE: (Notify the assistant head lower than the body. immediately) Put the bunsen burner off. Use shower in case of hair and clothes caches fire. Use the fire extinguisher when necessary. BLEEDING: compress on the wound, keep the wound CHEMICAL SPILL: clean in a above the heart level and get medical help. manner appropriate to the chemical. Aqueous solutions can be removed with water. Information your assistant. ACID BURNS: Use NaHCO3 solution CHEMICALS SPILLED IN THE EYE: The Eye is washed immediately BASE BURNS: Use Boric acid or Acetic acid solution with plenty of water for at least 15 minutes (use the eye-wash shower rooms) Get medical help.

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Laboratory Security Pictograms According To the United Nation’s Globally Harmonized System of Classification and Labeling of Chemicals (GHS) Pictogram

Usage E (Explosive): Spark, heat, flame, pressure, or friction can cause explosion (R1R3). Keep away from spark, flame, heat, or pressure Protective dress should be worn and a safe distance should be kept.

F (Flammable): F+ (Extremely Flammable): Flammable and flashing (R10R12). Flashing point is below 35 °C. Keep away from open fires, sources of heat and sparks. O (oxidazing): Oxidizing substances can ignite flammable and combustible material or worsen existing fire and thus make fire fighting more difficult. Caution! Keep away from flammable, combustible and spontaneously combustible materials. (R7-R9). Protective dress should be worn and a safe distance should be kept. G (Gas): Compressed, liquidified, refrigerated, or dissolved gas. May cause its container to explode if heated or dropped.

C (Corrosive): Causes severe eye and skin irritation upon contact. Prolonged contact can causes severe tissue damage. May be harmful if inhaled. Do not breathe vapors; avoid contact with skin and eyes. (R34, R35).

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T (Toxic): T+ (Very Toxic-): Toxic (R23-R-25) and very poisonous (R26-R28). Very hazardous to health when inhaled, swallowed or when they come in contact with the skin. May even lead to death. Avoid contact with the human body and immediately contact a physician in case of contact. Xi (Irritant): Xn (Sensitizing): Causes severe eye and skin irritation upon contact. May be harmful if inhaled. Do not breathe vapors; avoid contact with skin and eyes. (R20-R22i R36-R38)

H (Healt effect): Sign for carcinogens, mutagens, teratogens, respiratory sensitizers and substances with target organ toxicity. Avoid contact with body/skin. Avoid inhalation. (R40, R45-R47). N (Toxic to environment): Hazards to the aquatic environment material. Must not be disposed or released to the environment.

All the chemical substances are identified according to the risk of hazard. The numbers and/or letters on the pictogram identify the hazard degree of the substance.

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LABORATORY EQUIPMENTS

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EXPERIMENT 1: IDENTIFICATION OF PURE SUBSTANCES BY THEIR PROPERTIES 1.1. Purpose Identification of the substances by utilizing the physical and chemical properties of them.

1.2. Theory Matter is anything that occupies and displays the properties of mass and inertia. All matter could be classified as pure substances and mixtures.

Pure substances: The composition and properties of an element or a compound are uniform throughout a given sample and from one sample to another. Element and compounds are called substances. Pure substances have characteristic chemical and physical properties. 1) Elements:

An element is a substance made from only one type of atom, that cannot be

broken down into simpler components by any non-nuclear chemical reaction

(C, Na, Mg,

N2). 2) Compounds: Chemical compounds are substances comprising atoms of two or more elements joined together. (H2O, AgCl, CaCO3).

Mixtures: 1) Heterogeneous mixtures: The composition and physical properties vary from one part of the mixture to another. (oil + water, sand + water). 2) Homogeneous mixtures: The mixtures is uniform in composition and properties throughout is said to be homogeneous mixtures (sugar + water, salt + water) A compound can be identified with the following characteristics:

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Figure 1.1. Classification of matter

Physical properties A physical property is one that a sample of matter displays without changing its composition. Melting point, boiling point, density, solubility, physical state, colour, and crystal structure can be given as an example of physical properties. In physical changes, appearance of the substance changes, but the composition of it doesn’t changes. For example; - Melting of ice - Dissolution of salt in water - Melting of cupper at high temperatures - Melting of candle are physical changes.

Chemical properties In a chemical changes the composition of matters changes and new kids of matters occur. According to this chemical properties is the ability of a substance to a variation in the composition under certain conditions. Chemical reactions are understood from the following observations: -Color change - Precipitation formation - Gas release - Heat and sound formation 17

For example; - Burning of sugar, - Making cheese from milk - Combustion of wood are chemical changes.

1.3 Materials and Chemicals Materials

Chemicals

Glass stick Test tubes Bunsen burner Wooden tongs Litmus paper (blue and red) Tube holder.

Salt (NaCl) Sugar (C6H12O6) Okzalic asid (H2C2O4) Sand (SiO2) Hypo (Na2S2O3) Lime (CaO) Cupper sulfate (CuSO4)

1.4. Experimental Procedure Tests given below is performed for all of these substances: 1. Heat change 2. Solubility in water 3. Acidity and basicity of the water solutions

1) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and heat on bunsen burner. * Observe the changes in the substance carefully and write down the type of the changes. 2) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and fill half of the test tube with pure water. Stir with glass stick and observe. Is the solution homogeneous, heterogeneous, is there any suspension formation? Write down your observations. If the solution is not homogeneous, heat gently and note if there is a change or not. Keep the solutions to use in part 3. 3) Put a drop of the solutions obtained in part 2 on a litmus paper using glass stick. Note if there is a color change. If RED litmus turns into BLUE

→ compound is BASE 18

If BLUE litmus paper turns into RED → compound is ACID If there is no change in the color

→ compound is NEUTRAL

4) Ask your assistant for your unknown substance. Perform the testes done above to identify the substance. (Unknown substances: All, except sand, cupper sulfate and sugar) *All the test tubes used should be clean. Any contamination may cause you to find wrong compound.

Cautions: Wear protector glasses to protect your eyes. Don’t smell released gasses.

1.5. Results and discussion Heating

Type change

of Solubility in Acidity/Basicity water

Salt (NaCl) Hypo (Na2S2O3) Oxalic asid (H2C2O4) Cupper (II) sulfate (CuSO4) Sugar (C6H12O6) Lime (CaO) Sand (SiO2) Unknown No: Formula:

1.6. Questions 1. What is chemical and physical change? Explain and give an example for each. 2. What are the observations to identify chemical change? 3. How do we identify whether a compound is acidic or basic? 4. Is heat always cause same change? Think about your observations for the answer

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EXPERIMENT 2: IDENTIFICATION OF SUBSTANCES BY THEIR REACTIONS

2.1. Purpose Investigation of the reactions given by the compounds

2.2. Theory Chemical reaction is change of one or more pure substances into new substances as a result of electron mobility causing bond breakage /formation under specific conditions. There are many types of reactions and can be classified in different ways.

Chemical reactions may be grouped in four main groups:

1. Coupling reactions: A compound forms by combination of the two or more elements. A + B → AB CO2 + H2O→ H2CO3 2. Decomposition reactions: The reaction is the opposite of the coupling reaction and degradation of a compound of two or more simple substances. AB → A + B NaCl → Na+ + Cl-

3. Single displacement reaction: A reaction that an element is displaced to the other element in a compound. A + BC → AC + B Mg + 2HCl → MgCl2 + H2 4. Double substitution reactions: A reaction that displacement takes place between plus or minus charged ions in the reaction. Such reactions can take place only in solution and may end with, a) Precipitation formation

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b) Gas release

Positive charge exchange ↓͞ ͞ ͞ ↓ AB + CD → AD + CB KCl(aq) + AgNO3(aq) → KNO3(aq) + AgCl(k) K2CO3(aq) + HCl (aq) → KCl(aq) + H2O(s) + CO2(g) Precipitation formation is defined by the solubility rules given below.

General Rules for the solubility of ionic compounds in water.

Water Soluble Compounds Na+, K+ and NH4+ salts NO3 -, C2H3O2 - and ClO3 - salts Cl -, (Br -) and (I -) salts, (except Ag+, Pb+2 and Hg2+2 compounds) SO4 -2 salts (except Ba+2, Pb+2 and Sr+2 compounds) CO3 -2 and PO4 -3 salts (except Na+, K+ and NH4+ compounds) OH- and O-2 salts (except Na+, K+, NH4+ and Ba+2 compounds) S-2 salts ( except Na+, K+, NH 4+ and Ba+2 compounds)

Compounds which are not soluble in water: If a compound is not soluble in water, can be made soluble by the methods below. 1) pH of the solution can be adjusted. 2) Oxidation- reduction reactions can be used. 3) Water- soluble complex of the compound can be formed.

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2.3 Materials and Chemicals Materials Test tubes (6-8) Bunsen burner Wooden tongs Tube holder. Water bottle Baget

Chemicals Sodium carbonate (Na2CO3) Magnesium oxide (MgO) Sodium sulfate (Na2SO4) Magnesium sulfate (MgSO4) Lead nitrate(Pb(NO3)2) Barium chloride (BaCl2) Barium nitrate solution (BaNO3)2(aq) Hydrochloric acid solution HCI(aq) Sulfuric acid solution H2SO4(aq), Nitric acid solution HNO3(aq)

2.4. Experimental Procedure List of the compounds are given below: Cupper sulfate (CuSO4) Ammonium chloride (NH4Cl) Sodium carbonate (Na2CO3) Sodium sulfate (Na2SO4) Magnesium oxide (MgO) Magnesium sulfate (MgSO4) Lead nitrate (Pb(NO3)2) Barium chloride (BaCl2) Tests given below is performed for all of these substances: 1) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and heat on bunsen burner. * Observe the changes in the substance carefully and write down the type of the changes. 2) a) Find the solubility of each substances in water by using solubility rules and save in your report as soluble or insoluble b) Prepare the solutions of the compounds soluble in water. Put a very small amount of the substance in a test tube and fill half of the tube with water. Dissolve the compound by stirring with glass stick. c) Divide the homogeneous solutions obtained into 3 parts. Add 22

* 2-3 drops of Ba(NO3)2 (Barium nitrate) solutions, into first test tube * 2-3 drops of HCl (Hydrochloric acid) solutions, into second test tube * 2-3 drops of H2SO4 (Sulfuric acid) solutions, into third test tube Note any change. If there is a change (as precipitation, color change, gas release, suspension formation) write down the reaction for each substance. 3) a) If the substance is not soluble in water. Dissolve it in HNO3 (Nitric acid) solution. Note if there is a change or not. b) repeat the test in 2-c 4) Ask your assistant for your unknown substance. Perform the testes above to identify the substance. (Unknown substances: All, except cupper sulfate) (NOTE: All the test tubes used should be clean. Any contamination may cause you to find wrong compound)

Cautions: Wear protector glasses to protect your eyes.

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2.5. Results and discussion

Heating

Solubility H2O/HNO3

No:

Formula:

Ba(NO3) addition

HCl addition

H2SO4 addition

Sodium carbonate (Na2CO3) Sodium sulfate (Na2SO4)

Lead nitrate (Pb(NO3)2)

Cupper sulfate (CuSO4)

Ammonium chloride(NH4Cl)

Magnezium oxide(MgO)

Barium chloride(BaCl2)

Unknown

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2.6. Questions 1. Write down the type of reactions below. a) Fe3+ + 3OH− → Fe(OH)3________________ b) NaClO3→ NaCl + 3/2 O2 ________________ c) Zn+ 2HCl → ZnCl2 + H2 ________________ d) 2NaOH + CaCl2 → Ca(OH)2 + 2NaCl ________________

2. Give the compound name for the following ; a) Dissolves NaCl but not the AgCl ________________ b) Dissolves Na2CO3 but not the BaSO4’ı ________________ c) Dissolves both ZnS and Ag2S ________________ d) Dissolves both CaO and Sr(OH)2 ________________ 3. Give the compound name for the following ; a) Forms precipitate both with NaCl and FeCl2 ________________ b) Forms precipitate both with KCl and K2CO3 ile ________________ c) Forms precipitate with AgNO3 but not with Zn(NO3)2 ________________ 4. Complete the possible ones in the reactions below a) CaCO3 (k) + HNO3 (sulu) → b) CuBr2 (aq) + K2S (aq) → c) Sr (s) + O2 (g) → d) KCl (aq) + AgNO3 (aq) → e) BaSO4 (s) + NaCl(aq) →

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EXPERIMENT 3: STOCHIOMETRIC CALCULATIONS 3.1. Purpose Determination of the reaction stoichiometry, calculation of the theoretical and actual yield.

3.2. Theory Many information like mol ratios and physical states of the reactants and products can be obtained from a balanced reaction equation. For example; 2Cu(NO3)2 (s) → 2Cu (aq) + 4NO3-(aq) Ba(NO3)2 (aq) + K2SO4 (aq) → BaSO4(s) + 2KNO3 (aq)

In all the reactions, reactant compounds combine in specific ratios. If one of the reactants is used in excess, excess amount remains in the reaction medium without reacting. Completely used reactant in a chemical reaction is called limiting component and this component determines the amount of products. Amount of a product calculated from the balanced equation of a chemical reaction is called as theoretical yield. Many reactions are completed, in such cases the amount obtained is always less than theoretical yield. Amount obtained at the end of a reaction is called real yield.

Percentage yield of the reaction is calculated using the following formula:

Real Yield % yield =

x 100 Theoretical yield

In this experiment the following reaction is going to be performed and percentage yield will be calculated.

FeCl3(aq) + 3K2C2O4.H2O (aq) → K3Fe(C2O4)3.3H2O(s) + 3KCl(aq)

(Iron(III) chloride + potassium okzalate → potassium ferritriokzalate trihidrate + potassium 26

chloride).

3.3 Materials and Chemicals Materials 100 ml beaker 100 ml erlen mayer Funnel filter paper glass rod bunsen burner balance desiccator

Chemicals FeCl3(aq) K2C2O4.H2O (aq)

3.4. Experimental Procedure Synthesis of K3Fe(C2O4)3.3H2O salt 1. Weigh 4 gr. K2C2O4.H2O into a 100 ml. beaker and add 8 ml distilled water. 2. Heat gently (not be boiled) to dissolve the compound. 3. Add 4 ml FeCl3 solution into the hot solution by using a measuring cylinder . Color of the solution will turn to green. (1 ml solution contains 0,4 g FeCl3) 4. Solubility of the substance to be synthesized increases by temperature. Therefore, leave the beaker in ice bath for a while to precipitate the compound. 5. Filter the content of beaker by using funnel and weighed filter paper. Discard the decantate. Wash the crystals remained on the filter paper using a little amount of water. 6. Take the filter paper carefully from the funnel and place on a watch glass. (Don’t forget to write your name on the watch glass.) 7. Weigh the synthesized compound next week and use in the calculations.

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3.5. Results and discussion

FeCl3 solution volume : Filter paper mass: Filter paper + product mass Product mass FeCl3 mass: FeCl3 mole: K2C2O4.H2O mass : K2C2O4.H2O mole: Excess reactant mass and mole: Product mole (obtained): P roduct mole ( theoretical): % Yield:

3.6. Questions 1. In the precipitation reactions, obtained product mass is always less than the expected one. Explain, why? 2. Reaction of 1.5 gr. CaCl2 and 3.4 gr. AgNO3 gives 2.5 gr. AgCl. Write the equation for this reaction and calculate the % yield. (CaCl2: 111 gr/mol, AgNO3: 170 gr/mol, AgCl: 144 gr/mol) 3. 3 ml, 0.1 M Pb (NO3)2 reacts with 12 ml 0.1 M K2CrO4. Calculate the amount of the precipitate (PbCrO4) and the concentrations of the ions remained in the solution. (PbCrO4: 323 g/mol)

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EXPERIMENT 4: ACID - BASE TITRATION 4.1. Purpose Preparation of acid and base solutions, learning of neutralization reaction.

4.2.Theory We encounter the concept of acids and bases in our daily life. For example, acid rain phenomenon is known as a current environmental issue. Acids and bases are one of the most important topics in chemistry since a majority of the chemical reactions are the acid and base reaction. Despite the wide variety of definitions of acids and bases, Arrhenius, Bronsted-Lowry and Lewis definitions are used today. Arrhenius Acid-Base Definition: Acid: Acid turns blue litmus paper red; is active with some metals (active metals) exposing hydrogen gas; has sour taste and forms hydrogen ions in the aqueous solution.

HCl HNO3 H2SO4

H2O

H+ + Cl-

H2O

H + + NO3-

H2O

2H+ + SO4-2

Base: Base turns red litmus paper blue and neutralizes acid. It forms OH- ions in aqueous solution;. Bases form water and carbonated compounds reacting with CO2. They don’t react with many of metal (except amphoteric metals).

NaOH Ca(OH)2 Al(OH) 3

H2O

Na++ OH-

H2O

Ca+2 + 2OH-

H2O

Al+3 + 3OH-

Lowry-Bronsted Acid-Base Definition:

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Acids are substances which give proton, bases are substances which take proton. According to scientists conjugate base is formed when acid gives proton and similarly conjugate acid is formed when base takes proton. Asit

Proton + Baz

Baz + Proton NH3 + H2O Baz (1) Asit (2)

Asit NH4+ + OHAsit (1) Baz (2)

Konjüge Asit Konjüge Baz

Lewis Acid-Base Definition: Acid is a substance which takes an electron or accepts electron pair. Base is a substance which gives electron or carries an electron pair. B: + A

BA

A = Lewis Asiti B: = Lewis Bazi

F

H H

N: H

Lewis Bazi

+

B

F

F

H

H

F

N

B

H

F

F

Lewis Asiti

If acid gives only one proton, it is named as monoprotic, mono functional, mono basic, mono equivalent. According to ionization capability in their aqueous solutions, acids and bases are classfied as ; a. Strong b. Mild c. Very mild

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a) Strong Acids and Bases: Acids or bases which are 100% ionizable in aqueous solution is called "strong acid or base". It is accepted that H3O+ ions completely originated from acid in strong acid solutions and OH- ions completely originated from bases in strong base solutions.

HCl + H2O

H3O+ + Cl-

NaOH H2O Na+ + OHb) Weak Acids and Bases (1): Weak acids or bases cannot ionized 100% in their aqueous solutions. Ionization of acids and bases is a chemical equilibrium reaction either with acidity constant, Ka or basicity constant, Kb.

CH3COOH + H2O NH3 + H2O

Ka Kb

H3O+ + CH3COO- Ka = 1,8. 10-5 NH4+ + OH-

Kb = 1,8. 10-5

c) Weak Acids and Bases (2): Weak acids or bases can partially ionized in their aqueous solution. Their Ka and Kb constants are very low whereas pKa and pKb are very high. H H3C

C

OH

Ka = 1,3. 10-16

H 2N

pKb = 9,13

H

Volumetric Analysis: Analysis of the synthesized or natural samples is one of the fundamental subjects of the chemistry. In chemical analysis, identification of the analytes in a sample called ‘’qualitative analysis’’ and determination the amounts of the analytes in a sample called ‘’qualitative analysis’’. There are two basic methods for quantitative analysis of the chemical samples in which no instruments are used. These methods are “gravimetric analysis” and “volumetric analysis”. In gravimetric analysis, amount of a known (identified) analyte is determined by measurement of mass. A water insoluble compound of the analyte is prepared and mass of the compound is used to calculate the amount of this analyte. Volumetric analysis is based on the measurement of the volume of a solution prepared in exact concentration of a substance which gives a complete reaction with the analyte. By using the measured volume, the concentration or mass of the substance can be calculated.

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Figure 4.1. Titration setup

Titration, also known as titrimetry, is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurements play a key role in titration, it is also known as volumetric analysis. In titrations, known concentration and volume of titrant reacts with a solution of analyte to determine the concentration . Standard solutions are used as titrant to determine the unknown concentrations of substances. Standard solution is a solution containing a precisely known concentration of an element or a substance. Chemical substances generally contain impurities, or therefore solution concentration can not be known precisely. Titrant solutions are standardized by titration of primary standard (pure, high molecular weight,..) substances. Volume of titrant used to reach the equivalent point is called titration volume. The equivalence point, or stoichiometric point, of a chemical reaction is the point at which chemically equivalent quantities of acid and base have been mixed. In other words, the moles of acid are equivalent to the moles of base. It can be found by means of an indicator, most often phenolphthalein. In a reaction, the equivalence of the reactants as well as products is conserved. The endpoint refers to the point at which the indicator changes colour in a colourimetric titration. Indicators are substances which show whether the substance tested is

32

basic or acidic. pH indicator is a chemical detector for protons in acid-base titrations. The reagent of known concentration and volume used in titrations. An acid–base reaction is a chemical reaction that occurs between an acid and a base. It is shown as; H+(aq) + OH-(aq)

H2O(s)

Ksu= 10-14

Reaction is balanced. If the reacting acid and base are both too strong, pH would be 7 at the time that the titration was completed. · Equivalence factor: The number of hydrogen and hydroxide ions that can be substituted for the acids and bases (e). · Equivalent weight: The ratio of the molecular weight to equivalence factor (E). · Amount of equivalent gram: The ratio of the grams of material to equivalent weight (EA). · Normality: Normality is a measure of concentration equal to the gram equivalent weight per liter of solution (N). Volume of solution: liter (V). Amount of dissolved matter: gram (m). Molecular weight: g/mol (MA) These terms using;

According to principle of equivalence; Na . Va = Nb . Vb

Students will find the amount of acid to neutralize 25 mL of 1 M sodium hydroxide. Acid and base will react to form sodium chloride which can be crystallized further. Neutralization reaction ; HCl + NaOH  NaCl + H2O

33

4.3 Materials and Chemicals Materials

Chemicals

Erlenmeyer flask (250 mL)

HCl solution

Burette fastener

1 M NaOH solution

Burette (50 mL or 100 mL)

Methyl orange (indicator)

Volumetric flask (25 mL or larger)

4.4. Experimental Procedure

Figure 4.2. Experimental set up for pH titration

In this experiment, sodium hydroxide forms sodium chloride-soluble salt with hydrochloric acid neutralization. This solution is concentrated and crystallized in the crystallization vessel.

1. Prepare 25 ml 1 M NaOH solution. 34

2. Mix pour 25 ml sodium hydroxide solution into an erlenmeyer flask and add two drops of methyl orange. 3. Full the burette with unknown molarity HCl solution end of the 0 point. 4. Add in a small volume of hydrochloric acid solution onto sodium hydroxide solution and rinse after each addition. Shake after any addition. Continue the addition of acid until a red solution appears. Please read through the burette volume of added acid. 5. Please read through the burette volume of added acid. 6. Repeat titration (4. step) two more times. And use the average value of these 3 titrations for your calculations.

Cautions: Wear protective gloves, gowns, glasses and a mask. Avoid breathing vapors.

4.5. Results and discussion HCl quantitaties which is needed neutralization. 1.Titration

2. Titration

3. Titration

Used NaOH volume (ml) Used HCl volume (ml) Normality of the HCl solution Normality of the NaOH solution Milliequivalent gram of used HCl Milliequivalent gram of used NaOH

4.6. Questions 1. Explain acids and bases definitions of Arrhenius, Bronsted-Lowry and Lewis. 2. Explain definitions of Titration, The equivalence point, Indicator, titrant briefly? 3. A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?

35

EXPERIMENT 5: DETERMINATION OF A SALT SOLUBILITY 5.1. Purpose Learning the general principles of solubility of salts.

5.2.Theory Solubility of a salt, generally in a certain amount of a solvent (typically 100 g) and at a given temperature, is calculated by taking into account how much salt is dissolved in a solution. Temperature is the main factor which influences the solubility. To increase or to reduce the temperature of a saturated solution changes the equilibrium and according to Le Chatelier's Principle, the direction of the equilibrium shifts to the direction of reducing changes. Increasing of temperature in the endothermic reaction, the equilibrium shifts to the endothermic direction; unlike decreasing of temperature shifts the equilibrium to the exothermic direction. Increasing the temperature increases the solubility of the most ionic compounds. Dissolution of more substances in a saturated solution is an endothermic process which requires extra energy (heat) and high temperature. It is more suitable to show changes in solubility versus temperature in a graph.

5.3 Materials and chemicals Materials

Chemicals

Mortar and pestle

NH4Cl or another salt

Test tube (20 cm)

Distilled water

Two-hole stopper Thermometer Copper wire Beaker (100 mL) Pasteur pipette

36

5.4. Experimental Procedure

Figure 5.1. Experimental set up for solubility of a salt

1. Choose one of the salts. 2. Prepare about 8 g of salt fine powder in a mortar. 3. Take a large test tube (20 cm) covered with an appropriate two-hole stopper. 4. Place a 110°C thermometer in one of the hole. Be careful that the thermometer must not touch the bottom of the tube. 5. Dip a copper-ring wire in the tube by using the another hole of stopper. Using the tip that the ring of copper wire should be able to move easily up and down manually. 6. Add 5 gr of the salt which crushed into fine powder and weight it again on analytical balance. 7. Add 3 ml of distilled water carefully by using a pipette and close the stopper on the tube. 8. Dip the tube into the boiling water in a 400ml beaker. Solution inside the tube should remain in the water bath, with a chelator attached spores and mixed by the ring bar until all the salt dissolved in the tube. Be careful not to break the thermometer in this process.

37

9. Add 1 ml of water, if the salt in the tube in the boiling water does not dissolve completely in a few minutes. 10. If necessary add 1 ml of distilled water again and continue until fully dissolved. Do not add too much water because it is very close to the desired degree of saturation of the solution. Do not hold the tube in boiling water more than necessary. Thus evaporative water loss is prevented in this way. 11. Remove the tube from hot water after all of the salt fully resolved. 12. Observe carefully the crystallization during cooling of the tube and determine temperature when the crystallization starts. This temperature is the saturation temperature of the solution. 13. Repeat this process several times to obtain an accurate saturation temperature in order to determine heating, cooling and crystallization temperature. 14. Add 1 ml of distilled water in the tube. 15. Keep the tube in the boiling water until all crystals dissolve again. 16. Remove the tube from hot water when the dissolution is completed. 17. Cool it down by applying continuous stirring. 18. Detect the temperature which crystals start to form. In this case repeat the heating and cooling process several times to obtain an accurate saturation temperature. Repeat this process (addition of 1 ml of pure) until obtaining at least 5 or 6 saturation temperature in a wide temperature range. So a different saturation temperature is obtained by adding 1 ml of distilled water. If addition of 1 mL of distilled water cause to a big change in saturation temperature, other experiments can be repeated by adding 0.5 ml of distilled water instead of 1 mL of distilled water. Conversely, if the addition of 1 ml of distilled water cause to a little change in saturation temperature, other experiments can be repeated by adding 2 or 3 ml of distilled water instead of 1 mL of distilled water.

Cautions: Wear protective gloves, gowns, glasses and a mask. Avoid breathing dust / smoke / gas / mist / vapor / spray. Keep away the explosives from the open flame and avoid releasing to the environment. .

38

5.5. Results and discussion Data for used salt: Experiment 1

Experiment 2

Experiment 3

Amount of the salt (g) Weight of test tube (g) Weight of test tube and salt (g) Weight of distilled water used to solve the salt (mL) Total weight of used distilled water (mL) Crystallization temperature (°C)

5.6. Questions

1. Please predict the solubility of your salt at some different temperatures.

2. Temperature (oC) / x-axis 92.1 70.1 49.6 31.2

Solubility (g Pb(NO3)2/100 g water) / y-axis 115 91.2 78.6 61.4

Draw a line between the points and beyond in both directions past the data points. Use this line to answer the following questions. a. What would the solubility be at 35 oC of lead nitrate? b. What would the solubility be at 25 oC of lead nitrate? c. What temperature would be necessary to make a solution with solubility of 65.0 g/100 mL?

39

EXPERIMENT 6: DETERMINATION QUANTITY OF Fe2+ WITH KMNO4 6.1. Purpose Learning the general principles of induction reduction reactions

6.2.Theory Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion. Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion. The total increase in the oxidation number must be equal to the total decrease in the oxidation number. S0 Oxidized (reducer)

+

O20



S4+O22-

Reduced (oxidizer)

When the oxidation number of sulfur atom was increased from 0 to +4, S was oxidized. When the oxidation number of oxygen atom was decreased from 0 to -2, O was reduced. Reduced agents are responsible for oxidation. Therefore this substance is called oxidizer. The two processes are interdependent so the oxidized agent is reducing. Two methods are commonly used for balancing of oxidation-reduction reactions. 1. The oxidation number method 2. Ion-electron method Redox titrations: Because of the large number of elements can be easily oxidized or reduced compared to other types of reactions, there are much more titration methods based on redox reactions. It is the most beautiful example the abundance and variety of material that can be assigned directly or indirectly with strong oxidation power and at the same time indicator feature showing the standard KMnO4 solution. Redox titrations similar to acid-base titration. Oxidizing and reducing agents react in equivalent amounts with each other. In equvalent value meg oxidizing substance=meg reducing substance (meg=miliequvalent gram). Knowing received or given electron numbers in redox reaction is required to calculate the equivalent weight of a compound used in a redox titration Example: KMnO4-  Mn2+ semi reaction 40

1 mol MnO4- takes 5 mol e-. Mn atom in permanganate ion has 7+ oxidation number and is reduced to Mn2+ ion by taking 5 e-. 1 mol KMnO4 is 158 g and combines with 5 mol e-. 1 mol e- is 31.6 g (1/5 mol). Generally, the equivalent weight of a compound used in a redox titration is found by the following equation. 𝐹𝑜𝑟𝑚𝑢𝑙𝑒 𝑤𝑒𝑖𝑔ℎ𝑡

Equvalent weight =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑝𝑒𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

6.3 Materials and Chemicals Materials

Chemicals

Burette

Sulfuric acid

Flask

KMnO4

Beaker

FeSO4 Pure water

6.4. Experimental Procedure

41

Figure 6.1. Experimental setup

Potassium permanganate is oxidizer and this effect changes depending on the acidic or alkaline environment. Usually KMnO4 titrations are performed in the acidic environment and in this case action valence of KMnO4 is 5. According to this, 1 liter 1/10 N KMnO4 solution has 1/50 mol/mol (3,16 g) KMnO4. 1. Solve 3.16 g KMnO4 in 1 L boiling water to prepare 0.1 N KMnO4 solution and boil 5 minute. Close the top and leave to rest for a night. Filter with glass fiber and keep in the bottle. 2. Take 250 ml clean, dry flask. Receive certain volume of fresh prepared Fe2SO4 solution. 3. Add distile water to be total volume 30 ml to solution. Add 10 ml 6N H2SO4 solution on. 4. Titrate with 0.1 N KMnO4 solution until the pink color seen. 5. Repeat the same process 2. and 3. times , if you have enough time. 6. Note your results in the table. Used reactions in this experiment: 2KMnO4 + 10 FeSO4 + 8H2SO4  2MnSO4 + 5Fe2(SO4)3 + 8H2O + K2SO4 MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 8H2O Semi reactions; Mn7+O4-  Mn2+ Fe2+  Fe3+ In this reaction Fe2+ ion is oxidized to Fe3+ by losing an electron. Equvalent weight of ferrous will be 55.85/molx1/1=55.85 equvalent gram.

Cautions: Wear protective gloves, lab coat, glasses and a mask. Avoid breathing vapors. Keep away from flammable materials and prevent spreading into the environment.

42

6.5. Results and discussion Determination Preucation of Quantity of Fe2+: 1. Experiment

2.Experiment

3.Experiment

Used FeSO4 Volume (ml) Used MnO4 Volume (ml) Normality of KMnO4 solution Normality of FeSO4 solution Used miliequvalent gram of KMnO4 Miliequvalent gram weight of Fe(II) Miligram quantities of Fe(II) in the sample Equvalent weight of Fe(II) (original value)

6.6. Questions 1. NH4+ , SO42-, H2O2, N2O, S2O32-, CN-, H3PO4, Cr2O72-, CrO42- Determine the oxidation state of ions groups and each atoms in the compounds. 2. Balance this reaction which accur in acidic solutions VO43- + Fe2+ 3. 5 SO32- + 2 MnO4- + 6 H+

VO2+ + Fe3+

5 SO42- + 2 Mn2+ + 3 H2O

What are the reduced and oxidized species in this redox reaction? 4. How to decide whether a chemical reaction is redox rection or not? 5. Balance the following redox reaction in basic solution. a. MnO2(k) + ClO3b. Fe(OH)3(k) + OClc. ClO2

MnO4- + ClFeO42- + Cl-

ClO3- + Cl-

43

EXPERIMENT 7: BUFFERS, BUFFER CAPACITY, AND BUFFERING ZONE 7.1. Purpose To learn how to prepare a buffer solution To determine buffering capacity of a buffer solution To determine buffering zone of a buffer solution

7.2.Theory Buffers are aqueous systems that resist changes in pH when small amounts of acid or base are added. Buffer solutions are composed of a weak acid (the proton donor) and its conjugate base (the proton acceptor). Buffering results from reaction equilibrium in a solution between a proton donor (HA) and its conjugate proton acceptor (A-).

The pH of any given buffer can be related to the pKa of the weak acid involved and the logarithm of the molar ratio of conjugate base to acid by means of the HendersonHasselbalch equation:

Stated more generally,

Whenever H+ or OH- is added to a buffer, the result is a small change in the ratio of the relative concentrations of the weak acid and its anion and thus a small change in pH. The decrease in concentration of one component of the system is balanced exactly by an increase in the other. The sum of the buffer components does not change, only their ratio.

OH- H2O A-

HA H+

Buffer capacity generally depends on the concentration of buffer solution. Buffers with higher concentrations offer higher buffering capacity. On the other hand, pH is dependent not 44

on the absolute concentrations of buffer components, but on their ratio. Buffer capacity is a term used to describe the ability of a given buffer to resist changes in pH on addition of acid or base. A buffer capacity of 1 is when 1 mol of acid or alkali is added to 1 liter of buffer and pH changes by 1 unit. Buffering Zone: Each conjugate acid-base pair has a characteristic pH zone in which it is an effective buffer. Since buffering is due to the equilibrium between HA and A-, this equilibrium is mostly established at pH values equal to pKa ± 1. Temperature coefficient (∆pKa/C) values: The pKa of a buffer, and hence the pH, changes slightly with temperature. Although the mathematical relationship of activity and temperature may be complicated, the actual change of pKa with temperature (∆pKa/C) is approximately linear. Table 1 presents the pKa and ∆pKa/°C for several selected buffers commonly used in biological experimentation.

Table 7.1. Some buffers used in biological assays.

Most biochemical experiments are influenced by pH, and it is therefore usually necessary to control the pH of any biochemical experimental system. A common way of accomplishing this is to perform experiments in buffered systems. Thus biochemical experimentation usually calls for buffer preparation. The purpose of this experiment is to illustrate buffer preparation by having you prepare an acetic acid-sodium acetate buffer of a desired pH. To prepare a buffer of a desired pH, it is necessary to prepare a solution containing the appropriate amounts of a weak acid and its conjugate base. 45

One way of preparing buffer is to mix the requisite amounts of the appropriate weak acid and its conjugate base. Another involves starting with the weak acid and generating its conjugate base by adding a strong base, or conversely starting with the conjugate base and generating the weak acid by adding a strong acid. If a pH meter is available, the appropriate mixture can be obtained by “trial and error” mixing of the solutions available for preparation of the buffer. However, if one does not have a pH meter or does not wish to use this approach, the appropriate amounts of the substances to be used can be found in biochemical handbooks or calculated from the Henderson-Hasselbalch equation. The latter approach will be used in this experiment.

7.3 Materials and Chemicals Materials

Chemicals

pH Meter buret,

1 M acetic acid solution

buret clamp, magnetic stirrer stirring bar

1 M sodium acetate solution

two ring stands utiliy clamp

1 M sodium hydroxide, NaOH, solution

250 mL beakers

distilled water

100 mL volumetric flasks glass pipets

7.4. Experimental Procedure A titration apparatus is built from the available stand materials together with a magnetic stirrer, beaker and burette (or pipette) (see Fig.1).

46

Figure 7.1. Experimental set-up for pH titration. Prepare a 0.1 M acetic acid-sodium acetate buffer having a pH of 5.0. Use HendersonHasselbalch equation to calculate the amounts of acetic acid and acetate ion contained in 100 mL of this buffer. 1) Prepare a 100 mL sample of this buffer by mixing the appropriate amounts of the 1 M acetic acid and 1 M sodium acetate solutions that will be provided. 2) Prepare a second 100 mL sample of the desired buffer by mixing the appropriate amounts of the acetic acid and 1 M sodium hydroxide solutions that will be provided. Note: in this case, be sure to start with the right amount of acetic acid. After preparing your two samples buffer, have your laboratory instructor check the pH of each. Keep the buffers you just prepared! Buffer capacity: The purpose of a buffer is to minimize changes in pH when either acids or bases are added to a solution. The ability of a buffer to resist pH changes depends both on the absolute concentration of the buffer components and the ratio of these to each other. At a given pH, a concentrated buffer will have greater buffering capacity than a diluted buffer. At a given concentration, the buffering capacity of a buffer will be greatest when the pH of the buffer equals pKa of the acid used to prepare the buffer. 1) After measuring the pH of your two buffer samples, combine them to give approximately 200 mL of your acetic acid-sodium acetate buffer. 2) Transfer 20 mL of this combined buffer to a clean 100 mL volumetric flask and make up to volume by adding water. Mix well. You now have two buffer solutions that differ five-fold in their concentrations. 3) Place a 25 mL sample of the undiluted buffer in a clean beaker. Measure the pH. Add 1 mL of 0.1 N NaOH. Mix well and measure the pH. Continue adding the NaOH in 1 mL increments and measuring the pH until the pH is approximately 1 pH unit above the pKa for acetic acid. At this point start adding the NaOH in 0.5 mL increments and 47

measuring the pH until the pH is approximately 2 pH units above the pKa for acetic acid (pKa for acetic acid = 4.75). 4) Place a 25 mL sample of the diluted buffer in a clean beaker. Measure the pH. Is it the same as the undiluted buffer? Add 0.5 mL of 0.1 N NaOH. Mix well and measure the pH. Continue adding the NaOH in 0.5 mL increments and measuring the pH until the pH is approximately 2 pH units above the pKa for acetic acid. 5) Place a 25 mL sample of the undiluted buffer in a clean beaker. Measure the pH. Add 1 mL of 0.1 N HCl. Mix well and measure the pH. Continue adding the HCl in 1 mL increments and measuring the pH until the pH is approximately 1 pH unit below the pKa for acetic acid. At this point, start adding the HCl in 0.5 mL increments and measuring the pH until the pH is approximately 2 pH units below the pKa for acetic acid. 6) Finally, place a sample of the diluted buffer in a clean beaker. Measure the pH. Add 0.5 mL of 0.1 N HCl. Mix well and measure the pH. Continue adding the HCl in 0.5 mL increments and measuring the pH until the pH is approximately 2 pH units below the pKa for acetic acid. Cautions: Acids and bases have a corrosive effect. Only small quantities and low concentrations are used In this experiment, therefore only a slight hazard exists. Wear protective glasses and a laboratory coat.

7.5. Results and discussion Data: Table 1. pH values of the buffers you prepared Buffer

pH

1 2 Table 2. Titration of undiluted buffer Volume of NaOH (0.1 N), mL

pH

Table 3. Titration of diluted buffer Volume of NaOH (0.1 N), mL

pH 48

Table 4. Titration of undiluted buffer Volume of HCl (0.1 N), mL

pH

Table 5. Titration of diluted buffer Volume of HCl (0.1 N), mL

pH

ADDITIONAL KNOWLEDGE Molarity and Ionic Strength of a buffer: In some of the biochemical literature, buffers are identified in terms of molarities, in others they are identified in terms ionic strengths. Accordingly, one needs to be familiar with these concepts. More important, however, one needs to be aware of the importance of the concept of ionic strength. The molarity of a buffer is defined as the sum of the molarities of the Brönsted acid and conjugate contained in the buffer. For example, in an acetic acid-sodium acetate buffer containing 0.045 moles of acetic acid and 0.055 moles of sodium acetate per liter the molarity would be 0.100. The concept of ionic strength differs from molarity in two ways. First, it considers only the ionic species present in solution. Secondly, it considers both the concentrations and the charges of these ionic species. Mathematically the ionic strength (µ) of a solution is defined as follows: 𝜇=

1 ∑𝑐𝑖 𝑍𝑖2 2

Where ci is the concentration of ion “i” and Zi is the charge of this ion. While µ is used in this experiment and much of the biochemical literature to denote ionic strength, frequently one will encounter the use of I for this quantity in the biochemical literature.

49

SAMPLE CALCULATION The ionic strength of 0.1 M NaCl can be calculated as follows: 1 ([𝑁𝑎+ ](1)2 + [𝐶𝑙 − ](−1)2 ) 2 1 𝜇 = [(0.1)(1) + (0.1)(1)] 2 1 μ = (0.1 + 0.1) = 0.1 2

𝜇=

In this case the ionic strength is numerically equal to the concentration of the solution. This will be true for all solutions of electrolytes composed of two univalent ions. The ionic strength of 0.1 M (NH4)2SO4 can be calculated as follows: 1 𝜇 = ([𝑁𝐻4 ](1)2 + [𝑆𝑂4 ](−2)2 ) 2 1 𝜇 = [(0.2)(1) + (0.1)(4)] 2 1 μ = (0.2 + 0.4) = 0.3 2 In this case the ionic strength is numerically larger than the concentration of the solution. This will be true for all solutions of electrolytes containing multivalent ions. The concept of ionic strength is important in systems that involve charged substances since the attractive and repulsive forces between these substances will be decreased as the ionic strength of the solution increases. For example, the ionization constant for acetic acid increases as the ionic strength of the solution increases. Of more obvious importance in biochemistry, are the interactions between charged solutes and the charged ion exchange matrices that are used in ion exchange chromatography, the intramolecular reactions between charged groups in proteins, and the intermolecular reactions proteins and charged substances. All of these interactions will be affected by changes in ionic strength.

SAMPLE CALCULATION Part I: Combining 1 M acetic acid and 1 M sodium acetate and water to obtain a buffer at a specified pH We will make use of two equations. This first is the following rearrangement of the Henderson-Hasselbach equation: 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼: 10

(𝑝𝐻−𝑝𝐾𝑎 )

[𝐴− ] (𝐴− ) = = [𝐻𝐴] (𝐻𝐴)

50

Equation I gives us the required ratio of concentrations (or, equivalently, of amounts) of conjugate acid and conjugate base (in terms of the pKa of the acid involved and the pH desired). The second equation we need is the following: 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼𝐼: [𝐻𝐴] + [𝐴− ] = 0.1 𝑀 Equation II simply states that the combined concentrations of conjugate acid (acetic acid) and conjugate base (acetate) must be 0.1 M (this is specified as part of the laboratory write-up). Notice that we have two equations in two unknowns, [HA] and [A-]. Now we simply solve for each unknown: From Equation I [𝐴− ] = [𝐻𝐴] × 10(𝑝𝐻−𝑝𝐾𝑎) Substituting for [A-] in Equation II, [𝐻𝐴] + [𝐻𝐴](10(𝑝𝐻−𝑝𝐾𝑎) ) = [𝐻𝐴] × (1 + 10(𝑝𝐻−𝑝𝐾𝑎) ) = 0.1𝑀 Rearranging, to solve for [HA], [𝐻𝐴] =

0.1𝑀 1 + 10(𝑝𝐻−𝑝𝐾𝑎)

And, of course, we have: [𝐴− ] = 0.1 𝑀 − [𝐻𝐴] Now, knowing (or being told) that the pKa of acetic acid is 4.76 and being asked to prepare a buffer of a particular pH (as an example, pH 4.06), we can easily figure what volumes of 1 M acetic acid and 1 M acetate we must use. [𝐻𝐴] =

0.1𝑀 0.1𝑀 0.1𝑀 = = = 0.083𝑀 (−0.7) 1 + 0.20 1.20 1 + 10 [𝐴− ] = 0.1 𝑀 − 0.083 𝑀 = 0.017 𝑀

These are concentrations in the final solution of 100 mL. It is simple, however, to calculate what volumes of the 1 M stock solutions must be used to give these final concentrations (with formulation N1.V1=N2.V2). Part II. Making a buffer at the same pH using 1 M acetic acid, 1 M NaOH, and water Once we realize that all of the acetate in this buffer solution will come from acetic acid itself (it is generated by adding NaOH to acetic acid), the calculations are essentially the same as performed in Part I. After NaOH and acetic acid and water have been combined in yet-to-bedetermined proportions, we must satisfy the same two equations: 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼: 10(𝑝𝐻−𝑝𝐾𝑎) =

[𝐴− ] (𝐴− ) = [𝐻𝐴] (𝐻𝐴)

and 51

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼𝐼: [𝐻𝐴] + [𝐴− ] = 0.1 𝑀 Thus, in the final solution, the ratio of base to acid must be given by Equation I and the concentration of acetic acid and acetate must sum to 0.1 M (as shown in Equation II). Since the acetate will come from whatever volume of acetic acid we chose to use, we must select 10 mL of 1 M acetic acid. (10 mL diluted to 100 mL will give a concentration of 0.1 M). So, we measure out 10 mL of 1 M acetic acid. Now the only remaining issue is to add an appropriate volume of 1 M NaOH to generate enough acetate so that the ratio given by Equation I is satisfied. But this is just the same ratio as in Part I, above. Hence, we add the same volume of 1 M NaOH as the volume of 1 M acetate used in Part I. Adding this volume of 1 M NaOH will generate just the right amount of actate and leave just enough acetic acid behind to give the ratio needed according to equation I.

7.6. Questions 1. Make a plot of pH (ordinate) vs. millimoles of base or acid added to the buffer (abscissa). Draw a smooth curve (best fit) through the points associated with each of your buffers (undiluted and diluted). 2. Which buffer is the better buffer? At what pH did your buffers show the greatest resistance to changes in pH? 3. What is the molarity of the buffer you prepared in the first part of this experiment? What is its ionic strength? 4. Outline how you could have prepared an acetic acid-sodium acetate buffer of the same pH but whose ionic strength is 0.1. 5. How many mL of 1 M KH2PO4 and 1 M K2HPO4 would you need to prepare a liter of 0.1 M phosphate buffer of pH 7? The pKa’s for phosphoric acid are 2.1, 7.2, and 11.8. 6. Show by balanced equations how each of the following buffers act when additions of HCl or NaOH are made: An acetic acid-sodium acetate buffer, a Tris hydrochloride-Tris buffer, a phosphate buffer of pH 7?

52

EXPERIMENT 8: POLYPROTIC ACIDS: DETERMINATION OF pKA VALUES USING pH TITRATION CURVES 8.1. Purpose To perform a titration with an acid. To determine the strength of an acid. To calculate the pKa value of an acid

8.2.Theory pH Titration Curves 'Idealized':To date the equivalence point of an acid base reaction has been determined using an indicator. In this experiment we are going to monitor the changes in pH that occurs during the titration of a weak polyprotic acid with a strong base. At the equivalence point one should expect to see a dramatic change in pH as the solution goes from acidic to strongly basic. Depicted on Figure 8.1. is an idealized pH titration curve for a weak diprotic acid. The first thing that you should notice is that there are two regions where we see a significant pH change. These, if you wish, correspond to two separate titrations. Titration 1 is the reaction of the first proton with the base (in this case sodium hydroxide). H2X(aq) + NaOH(aq) = NaHX(aq) + H2O (l) The second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na2X(aq) + H2O(l)

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Figure 8.1. Titration curve of weak polyprotic acid So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid. The overall reaction is the sum of the two titration's H2X(aq) + 2 NaOH = Na2X(aq) + 2 H2O(l) In determining the quantity of the acid or the molarity of the acid, we are normally just interested in the final equivalence point. In a pH titration plot, this is determined by finding the point of inflection on the final area where we see a significant rise in pH (This can be approximated by determining the midpoint.) However, this plot contains some other interesting features. First off, if we look at the area corresponding to the first titration, it should come as no surprise that its equivalence point corresponds to the addition of exactly 1/2 the volume of NaOH required to reach the final equivalence point. The real neat point comes at the 1/2 way point of each titration. Let us focus on the Titration 1. At the 1/2 way point, the concentration of H2X(aq) remaining in the solution is equal to 1/2 the initial concentration of H2X! The concentration of NaHX(aq) produced is also numerically equal to 1/2 the initial concentration of H2X! Let's focus for a moment on the acid equilibrium associated with the acid that we are dealing with in titration 1, H2X(aq) + H2O(l) = HX- + H3O+ Ka = [H3O+][HX-]/[H2X] or written in another way [H3O+] = Ka{[H2X]/[HX-]} using the concentrations that we know for H2X and HX- (=NaHX) at the 1/2 way point we get [H3O+] = Ka{1/2[H2X]initial/[1/2H2X]initial} [H3O+] = Ka 54

From the graph we can determine the pH at this point and since pH=-log10[H3O+], we can determine [H3O+] at this point and thus obtain the Ka for this equilibrium. Since this is a polyprotic acid, this corresponds to Ka1. Guess what you can determine from the pH at the midpoint of the second titration. This information can be used to help identify the acid in question since Ka for a large number of polyprotic acids are known. The first acid that you will be following today is citric acid which is an acid that falls into the idealized category. You should see three areas where the pH undergoes significant changes and should be able to determine the three Ka values for citric acid and compare the result to the three known values given below. H3C6H5O7(aq) + H2O(l) ↔ H2C6H7O7 - + H3O+ H2C6H5O7 - + H2O(l) ↔ HC6H6O7 2- + H3O+ HC6H5O7 2- + H2O(l) ↔ C6H5O7 3- + H3O+

Ka1 = 7.4x10-3 at 25oC Ka2 = 1.7x10-5 at 25oC Ka3 = 4.0x10-7 at 25oC

pH Titration Curves 'The Real World': In reality, many polyprotic acids only show one discernible equivalence point! The vast majority of the time, this corresponds to final equilibrium. If this is the case then, all the other equivalence points can be determined by knowing what type of polyprotic acid one is dealing with, i.e., diprotic or triprotic (Fig. 8.2). For a triprotic acid, the other two equivalence points should correspond to 1/3 and 2/3 of the volume of the base required to reach the final one and thus one can still determine the Ka values. The second acid that you will be looking at in this lab is an amino acid, glycine, whose Ka values are given in Fig. 8.2.

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Figure. 8.2. Conjugate acid-base pairs consist of a proton donor and a proton acceptor. Some compounds, such as acetic acid and ammonium ion, are monoprotic; they can give up only one proton. Others are diprotic (H2CO3 (carbonic acid) and glycine) or triprotic (H 3PO4 (phosphoric acid)). The dissociation reactions for each pair are shown where they occur along a pH gradient. The equilibrium or dissociation constant (Ka) and its negative logarithm, the pKa, are shown for each reaction.

8.3 Materials and Chemicals Materials buret, buret clamp, magnetic stirrer, stirring bar two ring stands, utility clamp pH Meter 250 mL beakers

Chemicals 0.02 M sodium hydroxide, NaOH, solution 0.02 M citric acid solution 0.02 M glycine solution distilled water

8.4. Experimental Procedure A titration apparatus is built from the available stand materials together with a magnetic stirrer, beaker and burette (Fig.2).

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Figure 8.3. Experimental set-up for the determination of the acid strength of citric acide. Your TA will also demonstrate the best set up for this experiment. 1. Using a graduated cylinder, place ~ 20 mL of the ~0.02 M citric acid into a small beaker. If necessary add distilled water such that the tip of the pH probe is covered. 2. Fill your buret with the ~0.02 M NaOH solution. Record the exact molarity of this solution. Record the initial buret reading. Remember that this corresponds to 0.00 mL of NaOH added. 3. Record the initial pH of the Citric acid. 4. Carefully add the NaOH recording the volume of NaOH required to effect a pH change of 0.2. Continue this process until the pH reaches 12. 5. Repeat steps one through four using the ~0.02 M glycine.

8.5. Results and discussion Data: Table 1. Your titration data: the volume of NaOH vs. pH Volume of NaOH (0.02 M), mL 0.00

pH (citric acid)

----

----

pH (gylicine)

---

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Note: There should be separate tables for citric acid and glycine titration data

8.6. Questions 1.

Plot a graph of 'pH' versus 'Volume of NaOH" added and from this graph determine: a) The Ka values for citric acid. b) The exact concentration of the citric acid.

2.

By making same things that is given above. a) Determine the Ka values for glycine. b) Find the exact concentration of the glycine.

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